What is the ph (to nearest 0.01 ph unit) of a solution prepared by mixing 58.0 ml of 0.0179 m naoh and 60.0 ml of 0.0294 m ba(oh)2?

-

12.59

Explanation: -

Strength of NaOH = 0.0179 M

Volume of NaOH = 58.0 mL = 58.0/1000 = 0.058 L

Number of moles = 0.0179 M x 0.058 L

= 1.04 x 10⁻³ mol

Mol of [OH⁻] given by NaOH = 1.04 x 10⁻³ mol

Strength of Ba(OH)₂ = 0.0294 M

Volume of Ba(OH)₂ = 60.0 mL = 60.0/1000 = 0.060 L

Number of moles = 0.0294 M x 0.060 L

= 1.76 x 10⁻³ mol

Mol of [OH⁻] given by Ba(OH)₂ =2 x 1.76 x 10⁻³ mol

Total [OH⁻] = 1.04 x 10⁻³ mol + 2 x 1.76 x 10⁻³ mol

= 4.56 x 10⁻³ mol

Total volume of the mixture = 58.0 + 60.0

= 118.0 mL

118.0 mL of the solution has 4.56 x 10⁻³ mol [OH⁻]

1000 mL of the solution has

= 0.0386 mol

Using the relation

pOH = - log [OH-]

= - log 0.0386

= 1.41

Using the relation

pH + pOH = 14

pH = 14 - 1.41

= 12.59

for points guys